Gas in reversible adiabatic expansion

Problem

A monatomic gas ($\gamma=5/3$) expands reversibly and adiabatically from volume $V_1=1\;\text{L}$ to $V_2=8\;\text{L}$. The initial temperature is $T_1=600\;\text{K}$. Calculate $T_2$ and $\Delta S$.

Solution

For a reversible adiabatic process $TV^{\gamma-1}=\text{const}$, from which $T_2=T_1\left(\frac{V_1}{V_2}\right)^{\gamma-1}$. With $\gamma-1=\tfrac{2}{3}$ and $\frac{V_1}{V_2}=\tfrac{1}{8}$: $T_2=T_1\left(\frac{V_1}{V_2}\right)^{\gamma-1}=600\left(\frac{1}{8}\right)^{2/3}=600\cdot\frac{1}{4}=150\;\text{K}$ A reversible adiabatic process exchanges no heat ($\dd Q=0$) and is reversible: it is therefore isentropic, so $\Delta S=0$. $\ev{T_2=150\;\text{K},\ \Delta S=0}$

Links

Topics: Entropy and the second law Concepts: Entropy · Thermodynamic processes · Second law of thermodynamics Skills: Entropy balance Objects: Ideal gas